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Find the centre and radius of the circles.$$x^{2}+y^{2}-8 x+10 y-12=0$$
$centre (4,-5) radius: \sqrt{53}$
Precalculus
Chapter 11
Conic Sections
Section 1
Introduction
Introduction to Conic Sections
Campbell University
Piedmont College
Oregon State University
University of Michigan - Ann Arbor
Lectures
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Determine the center and r…
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Find the center and the ra…
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in this problem we have the equation of the circle given us x is squared, yes Y squared -8 x plus 10. Y -12 equals to zero. This can Britain, nas, X square minus eight checks last 16 and this can be written as y square plus Ben, Y just 25. Now we see that are added 16 and 25 additionally, So I'll also subtract them 16 plus 25 is 41, So also subtract 41 -12 equals two zero. Now this becomes Explain us four whole square plus. Why plus five whole square equals to 41 plus 12 is nothing. But 53 on the standard equation of the circle is nothing but x minus h whole square glass. Why minus key? Whole Square equals two. R squared comparing these two. We have the center at it's common. Okay? Just nothing but four comma minus five and the radius. There's nothing. But under route 50, that's all.
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